🔒 Hey guys, day one of learning python and im trying to build a calculattor.

[urban hollow]

somehow, when i pick the option "sqrt" it does works but shows the second "else" too... how can i fix it?

#Trying to build a calculator

import math

n1=float(input('Pick a number:'))
opr=str(input('+, -, *, /, **, sqrt?'))
if opr == 'sqrt':
    s=math.sqrt(n1)
    print(f'The square root of {n1} is {s}')
else:
    n2=float(input('Another one:'))

if opr == '+':
    s=n1+n2
    print(f'The sum of {n1} and {n2} is equal  to {s}')
    
elif opr == '-':
    s=n1-n2
    print(f'The subtraction between {n1} and {n2} is equal to {s}')
    
elif opr == '*':
    s=n1*n2
    print(f'The product between {n1} and {n2} is equal to {s}')
    
elif opr == '/':
    s=n1/n2
    print(f'The division between {n1} and {n2} is equal to {s}')
    
elif opr == '**':
    s=n1**n2
    print(f'{n1} to the power of {n2} is equal to {s}')
    
else:
    print('Are you dumb? Please pick a valid operation/number!')

if i run (number) then sqrt, it shows:
Pick a number:
55
+, -, *, /, **, sqrt?
sqrt
The square root of 55.0 is 7.416198487095663 (first else)
Are you dumb? Please pick a valid operation/number (second else)

[sick spoke]

Hey @urban hollow!

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To do this, use the following method:
```py
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print('Hello, world!')```
You can **edit your original message** to correct your code block.

[sick spoke]

@urban hollow

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[slow kestrel]

you'll want to indent the entire second if-chain to be inside the first "else"

since you want it to only happen with the 2nd operand that you take there

[knotty vector]

if opr == 'sqrt':
    s=math.sqrt(n1)
    print(f'The square root of {n1} is {s}')
else:
    n2=float(input('Another one:'))

Regardless of which block you enter here, the next line is

if opr == '+':

you can also exit() in the if opr == 'sqrt': block.

[urban hollow]

Thanks guys! Love ya

[sick spoke]

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